₦airaland Forum

Hello clubkonnect, please how do i get 10GB SME data bundle on MTN instead of buying 5GB twice?

Pls how much is the fees?

Pls i want to apply but am currently in service, can i apply without NYSC certificate?

emmyeuler1:
Oml guys have finally gotten their payment....last Thursday.....strictly old awaerdees

Hello prof! Long time! Hw's studies? I want buy d form and i no know how 2 reach u oooooo

Am Sorry Mojeed4, i ain't a statistician;

given a set of N points chosen randomly in a plane, using any three of the points how many triangles can one obtain? There's a prize for the solution with proof.

LordPherule:
Help! Help!! Help!!!
Now, in the stage that Mathematics looks absurd. Kelebesgue measure indeed.
Pls provide me solutions to these:

i have been lukin around for a textbk dat treats dat but didnt find....can u suggest any, mayb i can help?

DatechMan:
Yes, I must admit there are cases where it is extremely close. But I don't think there exist a situation where it is actually 1.

There's an exception though. When the index is 1. e.g 1²⁵⁰

For the instances above:

3¹²⁹ has 62 digits.

129log3 = 61.5486... (Rounded up, we have 62) valid

3¹⁷ has 9 digits.

17log3 = 8.111... (Rounded up, we have 9) valid



Good Morning Nigeria

#GodBlessNigeria
[quote author=benji93 post=42432099]Oh isee thts ur idea of rounding up
and ofcourse not there is never a case where it is 0.1 or 1 exactly
it comes from the idea that it comes from the fact that every exponential expression a^b>1 is such that 10^n-1<a^b>10^n infact it was from this axiom i thought that a^b could be at any point at infinity very close to the lower boundary or the upper boundary, if it is very close to the upper boundary, where logk <1 normally based on your rounding up if you had n=999.62 forexample it would be 1000 but actually if the value of k is to considered you could have 999.62 + 0.4 which gives 1000.02 which should be approximated to 1001 somehow we would not be able to test all numbers to infinity, but we can test as many number of digits as possible, i am not trying to disprove you bro,i am just saying based on my assumption there is a possibilty of having that +-1 problem somewhere at infinity unless my assumption is wrong which is also another possibility...

i must commend u 4 dis work sir, its the most significant work on this thread 4 more than a year now, ur idea is good but i wish to make a few corrections, Basically the part u find difficult to answer; "when to add 1, subtract 1 or approx.", on that note, u should b aware that BENJI93 is also very much on track!!!


Let it be required to find the number of digits in d^c NOTE; the limitation of ur proof subconsciously assumes that "d" must be a unit digit. Here is a proof that removes that assumption
now d^c=d^cx10^c/10^c

=(d/10)^cx10^c

let (d/10)^c=u, then if d is a unit digit, ur inequality 0.1<d<1 is obviously satisfied, otherwise it fails. So lets remove ur UNCOUNSCIOUS assumption by letting "d" be an r-digit number. Then d=kx10^r-1 where 1<k<10, so u=(0.kx10^r-1)^c=0.k^cx10^(r-1)c

hence, d^c=0.k^cx10^(r-1)cx10^c=0.k^cx10^rc. Suppose we let m=0.k^c, then since m=10^logm we have; d^c=10^logmx10^rc=10^rc+logm, From this we conclude that the number of digits in d^c are precisely 1+rc+clog0.k [or 1+(r-1)c+clogk); if the first digit of the base number is 1 (one) then, there are precisely 1+(r-1)c digits and that is the only case where we know the result precisely (without using calculator). Finally, the number of digits of d^c is greater than 1+(r-1)c, where r is the number of digits in d.

For example, if we denote "N" by the number of digits in d^c, then;
for 11², N=1+1x2=3
for 100², N=1+2x2=5

HI BENBUKS!!!

[quote author=agentofchange1 post=41911781]Mathematicians, philosophers, and enthusiasts Let’s see what the masters of the field have to
say about mathematics. Don’t forget to type on
the comment box which of the quotes is/are
your favorite/s.
1. Mathematics is a game played according to
certain rules with meaningless marks on paper.
— David Hilbert
2. Mathematics is concerned only with the
enumeration and comparison of relations. — Carl
Friedrich Gauss
3. Mathematics is the door and key to the
sciences. — Roger Bacon
4. Mathematics is the science of what is clear
by itself. — Carl Jacobi
5. Mathematics – the unshaken Foundation of
Sciences, and the plentiful Fountain of
Advantage to human affairs. — Isaac Barrow
6. Mathematics is as much an aspect of culture
as it is a collection of algorithms. — Carl Boyer
7. Mathematics is the art of giving the same
name to different things.– Henri Poincaré
8. Mathematics is one of the essential
emanations of the human spirit — a think to be
valued in and for itself like art or poetry.
9. Mathematics is not a careful march down a
well-cleared highway, but a journey into a
strange wilderness, where the explorers often get
lost. Rigour should be a signal to the historian
that the maps have been made, and the real
explorers have gone elsewhere. – – W. S. A
10. Mathematics, as much as music or any other
art, is one of the means by which we rise to a
complete self-consciousness. The significance of
mathematics resides precisely in the fact that it
is an art; by informing us of the nature of our
own minds it informs us of much that depends
on our minds.– John William Navin Sullivan
11. Mathematics is the supreme judge; from its
decisions there is no appeal.–Tobias Dantzig
12. Mathematics is no more computation than
typing is literature.– John Allen Paulos
13. Mathematics is the language with which God
wrote the universe. — Galileo
14. Mathematics is the handwriting on the
human consciousness of the very spirit of life
itself. — Claude Bragdon
15. Mathematics is the queen of science. — Carl
Friedrich Gauss
16. Mathematics may be defined as the subject
in which we never know what we are talking
about, nor whether what we are saying is true. —
Bertrand Russell
17. Mathematics consists in proving the most
obvious thing in the least obvious way. — George
Polya
18. Mathematics is like love; a simple idea, but it
can get complicated.
19. Mathematics is like checkers in being
suitable for the young, not too difficult, amusing,
and without peril to the state. — Plato
20. Mathematics is not a deductive science –
that’s a cliché. When you try to prove a theorem,
you don’t just list the hypotheses, and then start
to reason. What you do is trial and error,
experimentation, guesswork. — Paul Halmos
21. Mathematics is written for mathematicians.
— Copernicus
22. Mathematics is a great motivator for all
humans.. Because its career starts with zero and
it never end (infinity).
23. Mathematics is not a book confined within a
cover and bound between brazen clasps, whose
contents it needs only patience to ransack; it is
not a mine, whose treasures may take long to
reduce into possession, but which fill only a
limited number of veins and lodes; it is not a
soil, whose fertility can be exhausted by the
yield of successive harvests; it is not a continent
or an ocean, whose area can be mapped out and
its contour defined: it is limitless as that space
which it finds too narrow for its aspirations; its
possibilities are as infinite as the worlds which
are forever crowding in and multiplying upon the
astronomer’s gaze. — J. Sylvester
24. Mathematics is, I believe, the chief source of
the belief in eternal and exact truth, as well as a
sensible intelligible world. — Bertrand Russell
25. Mathematics is the tool specially suited for
dealing with abstract concepts of any kind and
there is no limit to its power in this field . — P.
Dirac.
26. Mathematics is not only real, but it is the
only reality. — Martin Gardner
27. Mathematics is often erroneously referred to
as the science of common sense. — Newman &
Kasner
28. Mathematics is the cheapest science. Unlike
physics or chemistry, it does not require any
expensive equipment. All one needs for
mathematics is a pencil and paper.
29. Mathematics Is an Edifice, Not a Toolbox
30. Mathematics is an independent world
created out of pure intelligence. — William
Woods Worth
31. Mathematics is, as it were, a sensuous logic,
and relates to philosophy as do the arts, music,
and plastic art to poetry. — K. Shegel
32. Mathematics is a more powerful instrument
of knowledge than any other that has been
bequeathed to u
superb!!!! Nice work!!

stlaibrowne:
Good job though but I don't totally agree with the h (60) interpretation because at t =60 the height is negative doesn't necessarily means the rocket is back to ground level it might as well means the time function is not valid for state or height of rocket at t=60

ur interpretation is right sir...nice criticism!!!

ladokuntlad:
1)
h(10)=-5(10)²+100(10)
=-5(100)+1000
=500
implies that at t=10(tenth seconds), the rocket is 500feet in the air.
h(60)=0
because wen we insert dat into d equation we get a negative result.
and its not posible for a height to neative. dat implies dat at t=60 the rocket is already back to the ground and its at rest hence height is zero(0)
[/b]

actually h(60)=0 ONLY because of the stated boundary condition, not for ANY of the reasons u stated...have a nice day&kudos

ladokuntlad:
In naija arguing tone

Dis a situation wia google doesnt even have the answer

If e sure for u ansa dis ones.

I give una d whole of this year
cc:stlaibrowne, MathsChic, Karmanaut, Laplacian, madmatician, agentofchange1, masperano, bolkay47, Madmathecian, killsmith, Ayomide002.
NO MORE BABY MATHS
Note: strictly for graduates and post graduates students.

u might want 2 type dis questions!!!

ladokuntlad:
U missed number 1.
Its 162+171=333
U got number two though

there are only two heptagons meeting the conditions, sum of their respective angles are;
144+171=315 and
153+162=315

[quote author=MathsChic post=40051985][/quote]1.) the sum of the largest two angles in d heptagon is 315 degrees
2.) the largest number wit that property is 1863, when divided by 1000, the remainder is 863

Karmanaut:
Laplacian, welcome back.

tnx bro...along wit Benbuks, i luv ur efforts here so much; seems we 've lost most of our generals

MathsChic:
Problem 2
(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(-√2+√11+√13)
=-(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(√2-√11-√13)

=-(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(√2-√11-√13)

=-((√2+√11)²-(√13)²))((√2-√11)²-(√13)²)
=-(2+11+2√22-13)(2+11-2√22-13)
=-(2√22)(-2√22)
=4.22
=88

can u write out one or two of his other questns?

ladokuntlad:
my desert for today

too tiny

Antoinne:
you sure about this?

100%

Antoinne:
Good work. You on track. I was hoping you'd give the full solution, though. So, we can see the relationships clearly and see if it offers any insights (hyperbolic or parabolic trigonometric functions).
Good try still.

ok. I was in a car then.
Let k=√[1+(2a/n)²] then,
S=2aLog[(k+1)/(k-1)]

N=S/L=(1/#)Log[(k+1)/(k-1)].
An important deduction is that no circle can ever make whole number of revolutions

Antoinne:
So, I have a problem that occured to me while solving the parabola problem earlier. Don't know if you guys are interested

Given a parabola, a circle with radius the length of parabola focus is placed at a point N on the parabola, n distance away from the center of the parabola. If this circle falls along the parabola and rises to the other end of the parabola, equally n distance away to the right, how many revolutions will it make? You can express answer in terms of all constants.

I'm wondering this may have some application with planetary bodies

tags: agentofchange1, dejt4u, Laplacian, Antoinne, Karmanaut, doubleDx, etc

all things being equal, the equation of the parabola is;
4ay=x², now the length of any curve is given by;
S=§√[1+(y')²]dx, in our case, y'=x/2a, integrate from x=-n to x=n,
S=§√[1+(x/2a)²]dx, substitute x=(2a)tan@, the integration should be easy for you!! The length of the circle is
L=2#a, so the number of revolutions N=S/L.
Am lukin for a concise way of solving Jackpot's second problem without rotating the axis.

agentofchange1:
#just. drinking. Garri and observing


happy solving guys...

@ Laplacian ( m ), Karmanaut greetings in CAPITAL letters

i feel ur efforts bro, kudos!!...enjoy urself!!!

Antoinne:
Corrected. thanks

nice wrk!

Antoinne:
Parabola of x = (3-2y)(4y+3)
x = 12y + 9 - 8y² - 6y
x = 9 + 6y - 8y²

Parabola is about y, since y has two values.
We can treat this as a mirror y = 9 + 6x - 8x², and then invert the mirror at the solution, if x is easier to work with.

anyways, for a parabola y = ax² + bx + c with a focus at p distance above the vertex V(h,k).
dy/dx = 2ax + b. At the vertex, you have the least variation, so dy/dx=0
2ax + b = 0
x = -b/2a, this should correspond to h.

y = ax²+bx+c at x = -b/2a, k
k = (4ac - b²)/4a

Assuming parabola at origin and following definition y = 4px²
Move parabola to right of xy plane so that vertex is V(h,k)
4p(y-k) = (x-h)²
Relating coefficients produces p = 1/4a, (focus)

Latus-rectum is length across focus.
if Parabola at origin (0,0)
4py = x²
with y = p on curve
x² = 4p.p = 4p²
x = 2p

Length of latus-rectum
2.x = 4p

Now, from earlier equation y = 9 + 6x - 8x² (mirror)
a = -8, b = 6, c=9

focus, p = 1/4a = 1/(4*-8 )
=-1/32

Vertex, V(h,k)
h = -b/2a = -6/(2*-8 ) = 3/8
k = (4ac - b²)/4a = 81 / 8

Length of latus-rectum = 4p
=abs(4.(-1/32)) = 1 / 8

Directrix equation: y = 325/32, i.e. (1/32 + 81/8 )

We can now go ahead and mirror our solution by just changing all x to y, and y to x
i. Directrix equation: x = 325/32
ii. Focus: -1/32
iii. Vertex: V(81/8, 3/8 )
iv. Length of it's latus-rectum: 1/8

Assuming parabola at origin and following definition y = 4px²?

jackpot:
Well, unlike integers that are either even or odd, a function may be neither even nor odd, eg h(x)=sin x-cos x, x+x².
Also, functions may have non-integral exponent, eg h(x)= x^e=x^{2.718281828. . .} or p(x)= £ x, where £ is the Gamma function or q(x)=x^2/7.

Now, my question is: how are we sure that there are no neither-even-nor-odd functions f satisfying the equation:
f(x)-f(-x)= -1/x ?

tags: Laplacian, agentofchange1, Laplacian , agentofchange1, doubleDx, Antoinne, dejt4u, AlphaMaximus, Laplacian etc

# with functions that are neither even nor odd, one can often find each part seperately and add them together.
# functions in which the variables are in integral powers are often called algebraic functions, and the theory of functional equations is usually restricted to this class of functions...elliptic functions, theta functions, mathieu functions e.t.c are not elementary functions and exhibit very cmplx properties which are unfamiliar and when incorperated into this subject makes it irrelevant and uninteresting
# there is no claim as to the uniqueness of a solution to a functional equation, since there maybe other solutions depending on the available properties and method of solution used

jackpot:
Please help out

Solve the functional equation below for f(x):

1. f(x)-f(-x)= - 1/x

2. f(x)-f(-x)=-1/x, given that f is odd.

tags: agentofchange1, dejt4u, STENON, AmazingAngel, Antoinne, Laplacian, Karmanaut, doubleDx, etc

if f(x)-f(-x)=-1/x then f(x) can neither be an even function nor have an even part, otherwise we have f(-x)=f(x) or f(x)-f(-x)=0, plug into d given eqn we have 0=-1/x, takin inverse makes x undefined. If f(x) is odd then f(-x)=-f(x), hence 2f(x)=f(x)-f(-x)=-1/x

jackpot:
Find the inverse laplace transform of 1/(s^5+a^5)

tags: Laplacian, agentofchange1, doubleDx, dejt4u, Alphamaximus, jaryeh

let k_n=e^2#ni/5, n=1,...,5 where # is pi
then,
1/(s^5+a^5)=product1/(s-ak_n)....resolve using partial fraction, cover-up rule (all factors are distinct) e.t.c

Antoinne:
Come on! No, they will not be equal. Look at the diagram again. "A" will have to be

A.cos(phi) = dx. Not A = dx.

You see it?

u'r right, they form a right angled triangle...maybe i shoul just take a rest!

In spherical co-ordinate, with standard notation, is it not logical to expect d elementary area on d sphere to be;
dS=r²d@d¤?
Why is it necessary to resolve in the xy plane to have;
dS=r²sin@d@d¤.
Note; @ is theta, ¤ is phi

Antoinne:
I took it you'd understand my little explanations.
Let's get the first part sorted. Look at the attached pic.

After resolving the elemental distances, dx & dy and multiplying. Follow the extension down to y = 1-x at z=0, you'll see that the distance dx on the other side is not exactly parallel to the distance you just resolved. I marked it with "A". There's still some kind of triangle there, showing that you elemental distance is not representative of the surface.

If that's too difficult, look at the basic 2D triangle below it. If you took a small portion on the x-axis called dx. Do you think it'll be equal to the length marked "A" on that same triangle merely by tracing it up?

It's the same thing happening on that surface above. You get it?

yes, i think they will b equal. This new diagram luks beta

Antoinne:
I suppose the way you were doing it was to resolve dx on one hand, and dy on the other to the lines at y=0 & x=0 respectively, so that you obtained dSx = sqrt(5)dx & dSy = sqrt(5)dy

And you then did, dS = dSx.dSy = 5dxdy. if this is how you did it, then you are wrong.

you should understand that even though, you had rightly resolved the elemental distances, you can't multiply them on that surface, because the surface is also bounded by y = 1-x (at z=0) which means in that plane, x is still dependent on y. If you had the snapshot below (Fig 1) where the elemental distance dy will not be dependent on x anymore on that surface, then it would have made sense.

So, it's easier to use vectors. Cos what you want to do is simultaneously apply the relationships given by the curves while still taking their dependences into consideration. In this way, you'd almost be resolving the surface itself dS and not the elemental distances.

Look at Fig2, where the same surface has been resolved onto the xy plane. On that xy plane, the last coordinate that will make the area the same is (2,2,0). So, if you had an elemental area on that surface dS₁, you can then obtain the angle between that area and the surface with the coordinate (1,1,2). Note that that is a mirror image of the surface in question, but in a different quadrant.

So, the angle between the two surfaces will be
(-1.5i-1.5j).(-0.5i-0.5j+2k) = sqrt(4.5).sqrt(4.5).cos(phi)
cos(phi) = 1/3.

dS.cos(phi) = dS₁
dS = dS₁/cos(phi) = 3dS₁

dS₁ = dxdy

So, dS = 3dxdy.

This elemental area will be the same as the elemental area on the main surface (they are simply mirror images).

Note: you could have resolved to the other side (270-360deg), and you'd have the same answer

i honestly appreciate ur effort sir! u explained exactly what i did, i know u have explained why it is not proper to resolve distances and obtain elemental area on the surface but i cant understand ur explanation on account of ur diagrams, i cant fit them together, pls snap a neat copy......i felt even if y depend on x at z=0, it should stil not affect d result because altering d original elementary area dS will also alter its projection on the surface...

jackpot:
Take f(x,y,z)=2x+2y+z-2=0.

dS= sqrt[(f_x)²+(f_y)²+1]dxdy=sqrt[4+4+1]dxdy=sqrt[9]dxdy=3dxdy.


Formula reference: www.math.lsa.umich.edu/~glarose/classes/calcIII/web/16_6/

ur method luks pretty, but d link didnt giv me d result...does it have name so i can google?

Am having trouble with an example in; Advanced Engineering Mathematics, K.A stroud: page 759, example 2; i understand their method but on applying a different method (which looks logical), am getting a different answer (dS=5dxdy). So, using any other method apart from the one provided, find the element of Area dS on the triangle. See fig. Below
tags; jackpot, agentofchange1, doubleDx, dejt4u, Alphamaximus, jaryeh

jackpot:
I guessed as much.

I solved and I got that the bomb landed a whooping 1106.57m. That's like 1.1km

any reason to justify that long distance the bomb traveled from a close-to-earth height of 600m?

i got something close 2 dat (1095.44m), approx, 1.1km; i guess its due 2 d speed of d air craft, since different value are obtaimd for different values of d speed